Prove that there exists $a\in K$ and a unit vector $e \in H$ such that $(a,e)\leq (x,e)\leq (a,e)+\delta$ for all $x \in K$.

Question

Problem

Let $K \subset H$ be a compact subset of a Hilbert space $H$ with diameter $\delta:=\sup_{x,y\in K}\|x-y\|>0$. Prove there exists an $a \in K$ and and a unit vector $e$ such that $$(a,e)\leq (x,e)\leq (a,e)+\delta$$ for all $x \in K$.

Attempt

This is equivalent to proving the existence of $a\in K$ and a unit vector $e$ such that $0\leq (x-a,e)\leq \delta$. for all $x\in K$. By Cauchy Schwarz, we only ned to show that $0\leq (x-a,e)$.

Alternatively, we need a linear functional with norm $1$ and an $a \in K$ such that $f(a)\leq f(x) \leq f(a)+\delta$.

Answer 1

Let $z(\neq 0)\in H$ .Define $e=\dfrac{z}{||z||}$.

Define $f:H\to \Bbb R$ by $f(x)=\langle x,e \rangle $ which is a bounded linear functional on $H$.

Now consider $f|_K: K\to \Bbb R$ i.e. restriction of $f$ to $K$. Since $f$ is bounded(continuous) and every continuous function on a compact set attains its minimum so it has a minimum in $K$ .

Hence there exists $a\in K$ such that $|f(a)|\le | f(x)|$ for all $x\in K$

$\implies | \langle a,e \rangle |\le |\langle x,e \rangle |\implies \langle x-a,e \rangle \ge 0$